For the group $G=\mathrm{GL}_n(\mathbb{K})$ of invertible matrices over any field, there is the well-known determinant character $\det:G\to\mathbb{K}^\times$ which, well, maps any matrix to its determinant. It has the property that
\[ [G,G]=\mathrm{SL}_n(\mathbb{K})=\{ g\in G \mid \det(g)=1 \}, \]
where $[G,G]$ denotes the derived subgroup of $G$. In more geometric terms, $[G,G]$ is the vanishing of the regular function $\det-1$ on $G$. Because I find it rather cute, I will show that you can find a similar function for all linear, reductive groups.
Let now $G$ be an affine, connected, reductive algebraic group over an algebraically closed field $\mathbb{K}$. I will quote the following result from Armand Borel's book

*Linear Algebraic Groups*, where it is the Proposition in 14.2:*Let now $n:=\mathrm{exp}(Q)=\min\{k\in\mathbb N\mid \forall q\in Q: q^k=1\}$ be the exponent of $Q$ and $\pi:C\to C$ the map $c\mapsto c^n$. We embed $Q$ into $C\times S$ as $\{ (q,q^{-1}) \mid q\in Q \}$. The proposition tells us that there is an isomorphism $\alpha:G\to(C\times S)/Q$ and that its inverse is induced by the multiplication $\mu:C\times S\to G$. Furthermore, the group $Q\subseteq C\times S$ is contained in the kernel of the morphism $(\pi\times1):C\times S\to C\times C$, so it factors as a morphism $\beta:(C\times S)/Q\to C\times C$. If $\mu\colon C\times C\to C$ denotes multiplication, then $\mu\circ\beta\circ\alpha$ is a homomorphism of algebraic groups \[ \begin{array}{rcl} \delta: G = C\cdot S &\longrightarrow & C \\[1ex] c\cdot s &\longmapsto& c^n \end{array} \] It has the property that $S=\delta^{-1}(1)$. In the case of $G=\mathrm{GL}_n(\mathbb{K})$, we have $S=\mathrm{SL}_n(\mathbb{K})$ and $C=\mathbb{K}^\times\cdot\mathbb 1_n\cong\mathbb{K}^\times$ is the set of invertible scalar matrices. The finite group $Q$ consists of all scalar matrices $\mu\mathbb 1_n$ with $\mu$ some $n$-th root of unity. The morphism $\delta$ is the determinant in this case.***Proposition.**Let $C$ be the identity component of the center of $G$. Then, $C$ is a torus and $S:=[G,G]$ is semisimple. Furthermore, $G=C\cdot S$ and $Q:=S\cap C$ is finite.